由單圈圖生成的凱萊圖的廣義3 連通度

王燕娜 周波

(1. 廣東交通職業(yè)技術學院基礎教學部,廣州,510650?2. 華南師范大學數(shù)學科學學院,廣州,510631)

1 Introduction

As usual,we denote byV(G)andE(G)the vertex set and edge set of a graphG. For a connected graphGand an integerkwith 2≤k ≤|V(G)|,the generalizedkconnectivity of a graphGis defined as[2,3,8]

whereκG(S)is the maximum number?of edge disjoint treesT1,··· ,T?inGsuch thatV(Ti)∩V(Tj)=Sfor every pair of distinct integersi,jwith 1≤i,j ≤?. In the following,we call such?trees as?internally disjoint trees connectingSinG.

Ifk=2,thenκk(G)is the smallest value of the maximum number of pairwise vertex disjoint paths between any vertex pair ofG,which is the ordinary(vertex)connectivity ofG[1]. Ifk=|V(G)|,thenκk(G)is the maximum number of pairwise edge disjoint trees ofG[9,10]. The generalizedkconnectivity may be used to measure the reliability and security of a network in whichknodes are users and other nodes are switchers.

LetXbe a group with identitye, and lete/∈S ?X, whereSis closed under inversion. The Cayley graph Cay(X,S) is a graph with vertexXsuch thatgandhforg,h ∈Xare adjacent if and only ifh=gsfor somes ∈S. We say Cay(X,S) is a Cayley graph onXgenerated byS. Denote Sym(n)the group of all permutations on[n] ={1,··· ,n}. By(p1,··· ,pn),we denote the permutationσsuch thatσ(i) =pifori ∈[n], and by [i,j] with 1≤i

which swaps the objects at positionsiandj.

LetTbe a set of transpositions from[n]. The(transposition)graph ofT, denoted byG(T), is the graph with vertex set [n] such that, fori,j ∈[n], verticesiandjare adjacent if and only if [i,j]∈T.It is known that the Cayley graph Cay(Sym(n),T) is connected if and only ifG(T) is connected. IfG(T) is the star, then Cay(Sym(n),T) is called a star graph, denoted bySn. IfG(T) is the path, then Cay(Sym(n),T)is called a bubble sort graph,denoted byBn. IfG(T)～=Cn(thenvertex cycle),then Cay(Sym(n),T)is called a modified bubble sort graph,denoted byMBn. IfG(T)is a general tree,then we denote by Tnthe graph Cay(Sym(n),T). IfG(T)is a general unicyclic graph,then we denote by Unthe graph Cay(Sym(n),T). In this case,we also say that Unis generated byG(T).

The generalized 3 connectivities of various Cayley graphs,for example,the star graph,the bubble sort graph,the modified bubble sort graph and even Tnfor any general treeG(T)have been determined,see,e.g.,[7,6,11,12],to mention just a few.

In this article, we show that the generalized 3 connectivity of Unfor any general unicyclic graphG(T)isn ?1 forn ≥3. We use the techniques developed in the above papers,especially in[6].

2 Preliminaries

Forv ∈V(G), denote byNG(v) the set of neighbors ofvinG, and letδG(v) =|NG(v)| andNG[v]=NG(v)∪{v}. For a subsetS ?V(G),denote byG[S]the subgraph ofGinduced byS.

Forx,y ∈V(G),a path fromxtoyinGis called an(x,y) path. Forx ∈V(G)andY ?V(G),an(x,Y) path is an(x,y) path inGfor somey ∈Y,and any other vertex of the path(if exists any)are not in{x}∪Y.

Lemma 2.1([1]) LetGbe akconnected graph, and letx ∈V(G) andY ?V(G){x}with|Y|≥k. Then there arekinternally vertex disjoint(x,Y) paths whose terminal vertices are distinct inY.

Lemma 2.2([4])LetGbe a connected graph with minimum degreeδ. Thenκk(G)≤δfor 3≤k ≤|V(G)|. Furthermore,if there exist two adjacent vertices of degreeδinG,thenκk(G)≤δ ?1.

Lemma 2.3([4])LetGbe a connected graph withnvertices. Ifκ(G) = 4k+r, wherekandrare two integers withk ≥0 andr ∈{0,1,2,3},thenκ3(G)≥3k+. Moreover,the lower bound is sharp.

Lemma 2.4([7]) Forn ≥3,κ(Tn)=n ?1.

Lemma 2.5([6]) Forn ≥3,κ(MBn)=nandκ3(MBn)=n ?1.

Suppose thatn ≥3. Recall that Un= Cay(Sym(n),T) whenG(T) is unicyclic. Obviously, Unis ann! vertex regular graph of degreen. Suppose thatG(T) ?Cn. ThenG(T)has a leaf(a vertex of degree one),and we assume thatnis a leaf ofG(T)withn ?1 being its unique neighbor. Fori ∈[n],let Symi(n)denote the set of all permutations of[n]{i}. Let

Similarly, ifG(T) is a tree, then we assume thatnis a leaf ofG(T) withn ?1 being its unique neighbor,andV(Tn)can also be partitioned intoV1,··· ,Vnand～= Tn?1,where= Tn[Vi]fori ∈[n].

IfQ=v1···vris apathinagraphG, thenQ?denotes thepathvr···v1. Foredge disjoint treesT1,···,Ts,if thegraph with vertexsetV(Ti)andedgesetE(Ti)isatree,thenwe denote this tree byT1+···+Ts.

Lemma 2.6LetGbe a graph obtained from two vertex disjointkconnected graphsG1andG2by addingtedges betweenG1andG2such that any two edges are not adjacent. Ift ≥k ≥1,thenκ(G)≥k.

ProofLetv1andv2be any two vertices ofG. It suffices to show that there arekinternally vertex disjoint paths betweenv1andv2. Ifv1,v2∈V(G1) orv1,v2∈V(G2), this is obvious asGiiskconnected fori=1,2.

Assume thatv1∈V(G1)andv2∈V(G2). Denote thetedges betweenG1andG2byxiyifori=1,··· ,t. Note thatt ≥k. LetS={x1,··· ,xk}andS?={y1,··· ,yk}. Obviously,|S|=|S?|=k. AsG1iskconnected,we have by Lemma 2.1 that ifv1?∈S,there arekinternally vertex disjoint(v1,S) pathsL1,··· ,Lksuch thatxi ∈Lifori= 1,··· ,k. Ifv1∈S,sayv1=x1,then there exists a(v1,S) path of length zero. So, there arekinternally vertex disjoint (v1,S) pathsL1,··· ,Lksuch thatxi ∈Lifori= 1,··· ,k. Similarly, there arekinternally vertex disjoint (v2,S?) pathsQ1,··· ,Qksuch thatyi ∈Qifori=1,··· ,k. Now we can obtainkinternally vertex disjoint(v1,v2) paths:Li+xiyi+fori=1,··· ,k.

Remark 2.1Suppose that{i,j} ?[n]withn ≥3. Note that Tn[Vi ∪Vj]may be obtained fromandby adding (n ?2)! edges betweenand, in which any two edges are not adjacent. By Lemma 2.4,κ() =κ() =κ(Tn?1) =n ?2. As (n ?2)!≥n ?2, we haveκ(Tn[Vi ∪Vj])≥n ?2 by Lemma 2.6. On the other hand,the minimum degree of Tn[Vi ∪Vj]isn ?2,soκ(Tn[Vi ∪Vj])≤n ?2. Thus,κ(Tn[Vi ∪Vj])=n ?2,see[7].

3 Main result

We need some lemmas that are used in the proof.

Lemma 3.1κ(U4)=4.

ProofIf U4～=MB4,then by Lemma 2.5,we haveκ(U4) = 4. Suppose that U4?MB4. As the minimum degree of U4is 4,κ(U4)≤4. In the following,we show thatκ(U4)≥4.

Letxandybe any two vertices of U4. It suffices to show that there are 4 internally vertex disjoint paths betweenxandy.

Case 1xandylie in the same main part of U4.

Assume thatx,yare in. Note that～=MB3. By Lemma 2.5,κ(MB3) = 3. Thus there are 3 internally vertex disjoint (x,y) paths, sayL1,L2,L3, in. As U4[V(U4)V1] is connected, there is an(x′,y′) pathQin U4[V(U4)V1]. It thus follows thatL1,L2,L3,xx′+Q+y′yare 4 internally vertex disjoint(x,y) paths.

Case 2xandylie in different main parts of U4.

Assume thatxlies inandylies in U23.

Case 2.1x′∈V2andy′∈V1.

Note thatx= (3,4,2,1)or(4,3,2,1), andy= (3,4,1,2)or(4,3,1,2). Suppose without loss of generality thatx=(3,4,2,1). Ify=(3,4,1,2),thenQ1=xy,Q2=x(4,3,2,1)(4,3,1,2)y,

and

are 4 internally vertex disjoint(x,y) paths. Ify=(4,3,1,2),thenQ1=xx′y,Q2=xy′y,

and

are 4 internally vertex disjoint(x,y) paths.

Case 2.2x′∈V2andy′/∈V1,orx′/∈V2andy′∈V1.

Assume thatx′∈V2andy′/∈V1. Assume thaty′∈V3. Thenx= (3,4,2,1)or(4,3,2,1), andy= (1,4,3,2)or(4,1,3,2). Suppose without loss of generality thatx= (3,4,2,1). Ify= (1,4,3,2),then

are 4 internally vertex disjoint(x,y) paths. Ify=(4,1,3,2),then

are 4 internally vertex disjoint(x,y) paths.

Case 2.3x′,y′∈V3orx′,y′∈V4.

Assume thatx′,y′∈V3. Thenx= (2,4,3,1) or (4,2,3,1), andy= (1,4,3,2) or (4,1,3,2).Suppose without loss of generality thatx=(2,4,3,1). Ify=(1,4,3,2),then

are 4 internally vertex disjoint(x,y) paths. Ify=(4,1,3,2),then

are 4 internally vertex disjoint(x,y) paths.

Case 2.4x′∈V3andy′∈V4,orx′∈V4andy′∈V3.

Assume thatx′∈V3andy′∈V4. Thenx= (2,4,3,1) or (4,2,3,1), andy= (1,3,4,2) or(3,1,4,2). Suppose without loss of generality thatx=(2,4,3,1). Ify=(1,3,4,2),then

are 4 internally vertex disjoint(x,y) paths. Ify=(3,1,4,2),then

are 4 internally vertex disjoint(x,y) paths.

Thus there are 4 internally vertex disjoint paths betweenxandyin all cases. The result follows.

Lemma 3.2Forn ≥3,κ(Un)=n.

ProofBy Lemma 2.5,it is true forn=3. Suppose thatn ≥4. We prove the result by induction onn.

By Lemma 3.1,it is true forn=4. Suppose thatn ≥5 and it is true for Un?1.

If Un～=MBn, then by Lemma 2.5, we haveκ(Un) =n. Suppose that Un?MBn. As Unisnregular, we haveκ(Un)≤n. So it suffices to show thatκ(Un)≥n, or equivalently, for any two verticesxandyof Un,there areninternally vertex disjoint paths betweenxandy.

Case 1xandylie in the same main part of Un.

Assume thatx,ylie in. Note that～= Un?1. By the induction hypothesis,κ() =n?1,so there aren?1 internally vertex disjoint(x,y) paths in,sayL1,··· ,Ln?1. As Un[V(Un)V1] is connected, there is an (x′,y′) pathQin Un[V(Un)V1]. It thus follows thatL1,··· ,Ln?1andxx′+Q+y′yareninternally vertex disjoint(x,y) paths in Un.

Case 2xandylie in different main parts of Un.

Assume thatxlies inandylies in. Note that(n ?2)!≥n ?1 forn ≥5.

Case 2.1x′/∈V2andy′/∈V1.

We may choosen ?1 vertices,sayz1,··· ,zn?1,ofV1such that∈V2fori=1,··· ,n ?1. LetS={z1,··· ,zn?1}andS?={,··· ,}. Clearly,|S|=|S?|=n?1. By the induction hypothesis,κ()=n ?1,so by Lemma 2.1,there aren ?1 internally vertex disjoint(x,S) pathsL1,··· ,Ln?1insuch thatzi ∈Lifori= 1,··· ,n ?1. Similarly, there aren ?1 internally vertex disjoint(y,S?) pathsQ1,··· ,Qn?1in,such that∈Qifori=1,··· ,n ?1. As Un[V(Un)(V1∪V2)]is connected,there is an(x′,y′) pathQin Un[V(Un)(V1∪V2)]. ThenLi+zi+fori=1,··· ,n?1,andxx′+Q+y′yareninternally vertex disjoint(x,y) paths in Un.

Case 2.2x′∈V2andy′∈V1.

We may choosen ?2 vertices different fromy′, sayz1,··· ,zn?2, ofV1such that∈V2fori=1,··· ,n?2. LetS={z1,··· ,zn?2,y′}andS?={,··· ,,x′}. Clearly,|S|=|S?|=n?1.By the induction hypothesis,κ()=n?1,so by Lemma 2.1,there aren?1 internally vertex disjoint(x,S) pathsL1,··· ,Ln?1insuch thatzi ∈Lifori= 1,··· ,n ?2,andy′∈Ln?1. Similarly,there aren ?1 internally vertex disjoint (y,S?) pathsQ1,··· ,Qn?1in, such that∈Qifori=1,··· ,n?2,andx′∈Qn?1. ThenLi+zi+fori=1,··· ,n?2,xx′+andLn?1+y′yformninternally vertex disjoint(x,y) paths in Un.

Case 2.3x′∈V2andy′∈/V1,orx′∈/V2andy′∈V1.

Assume thatx′∈/V2andy′∈V1. We may choosen ?2 vertices,sayz1,··· ,zn?2,ofV1different fromy′such that∈V2fori=1,··· ,n ?2,and choose a vertex,sayw,ofV2such thatw′∈/V1. LetS={z1,···,zn?2,y′}andS?={··· ,,w}.Clearly,|S| =|S?|=n?1. By the induction hypothesis,κ()=n?1,sobyLemma2.1,therearen ?1internallyvertex disjoint(x,S) pathsL1,··· ,Ln?1insuch thatzi ∈Lifori=1,··· ,n ?2,andy′∈Ln?1. Similarly,there aren ?1 internally vertex disjoint(y,S?) pathsQ1,··· ,Qn?1insuch that∈Qifori=1,··· ,n?2,andw ∈Qn?1. As Un[V(Un)(V1∪V2)]is connected,there is an(x′,w′) pathQin Un[V(Un)(V1∪V2)].ThenLi++fori= 1,··· ,n ?2,Ln?1+y′yandxx′+Q+w′w+areninternally vertex disjoint(x,y) paths in Un.

The result follows by combining the above cases.

Now,we are ready to prove the main result.

Theorem 3.1Forn ≥3,κ3(Un)=n ?1.

ProofIt is true forn=3 by Lemma 2.5. Suppose thatn ≥4. We prove this statement by induction onn.

By Lemma 3.1,κ(U4) = 4. So, by Lemma 2.3,κ3(U4)≥3. Asκ(U4) is 4 regular, we haveκ3(U4)≤3 by Lemma 2.2. Thusκ3(U4)=3. That is,the statement is true ifn=4.

Suppose thatn ≥5 and the statement is true for Un?1. If Un～=MBn,then by Lemma 2.5,we haveκ3(Un)=n?1. Suppose that Un?MBn. As Unisnregular,we haveκ3(Un)≤n?1 by Lemma 2.2.So it suffices to show thatκ3(Un)≥n ?1. LetSbe an arbitrary vertex subset of Unwith|S| = 3,sayS={x,y,z}. Then it suffices to show that there aren ?1 internally disjoint trees connectingSin Un.

Case 1x,yandzlie in some common main part of Un.

Assume thatx,y,zlie in. By the induction hypothesis,κ3()=n ?2,so there aren ?2 internally disjoint trees connectingS,sayT1,··· ,Tn?2,in. As Un[V(Un)V1]is connected,there is a spanning treeTin Un[V(Un)V1]. ThusT1,··· ,Tn?2andT+x′x+y′y+z′zaren ?1 internally disjoint trees connectingSin Un.

Case 2x,yandzlie in two different main parts of Un.

Assume thatx,ylie in U1n?1andzlies in U2n?1. Recall thatκ() =n ?1. So there aren ?1 internally vertex disjoint(x,y) paths,sayL1,··· ,Ln?1,in. Choosen ?1 distinct verticesx1,··· ,xn?1fromL1,··· ,Ln?1such thatxi ∈V(Li)fori= 1,··· ,n ?1. Note that at most one of these paths has length one. If there is indeed such a path of length one,sayL1,then in this case,we choosex1=x. LetZ={,··· ,}. Obviously,|Z|=n ?1. Asκ(Un[V(Un)V1])≥n ?1,we have by Lemma 2.1 that there aren ?1 internally vertex disjoint(z,Z) pathsQ1,··· ,Qn?1in Un[V(Un)V1]such that∈Qifori= 1,··· ,n ?1. NowTi=Li++Q?

ifori= 1,··· ,n ?1 aren ?1 internally disjoint trees connectingSin Un.

Case 3x,yandzlie in three different main parts of Un.

Assume thatxlies in,ylies inandzlies in. Evidently,we have eitherx′/∈V2orx′/∈V3. Assume thatx′/∈V2. Recall thatκ(Un[V1∪V2])≥n ?1. So there aren ?1 internally vertex disjoint (x,y) paths, sayL1,··· ,Ln?1, in Un[V1∪V2]. Asx′/∈ V2, all neighbors ofxinL1,··· ,Ln?1are in neighbors,which are denoted byx1,··· ,xn?1. LetZ1={,··· ,}. Assume that 2 appears in thejth position ofx. Clearly,j ?=n ?1. Ifx[j,n ?1] is a neighbor ofx, then(x[j,n ?1])′∈V2, and thus|Z1∩V2| = 1. Ifx[j,n ?1] is not a neighbor ofx, then/∈V2fori=1,··· ,n ?1,and thus|Z1∩V2|=0. It thus follows that|Z1∩V2|≤1.

Assume that{··· ,}/∈V2. LetZ2={x′,,··· ,}. Clearly,|Z2| =n ?1. Asκ(Un[V(Un)(V1∪V2)])≥n ?1,by Lemma 2.1,there aren ?1 internally vertex disjoint(z,Z) pathsQ1,··· ,Qn?1in Un[V(Un)(V1∪V2)]such that∈Qifori=1,··· ,n ?2,andx′∈Qn?1. NowTi=Li++fori=1,··· ,n ?2,andQn?1+x′x+Ln?1formn ?1 internally disjoint trees connectingSin Un.

The result follows by combining the above cases.

Remark 3.1This note shows that the generalized 3 connectivity of the Cayley graph Unon symmetric group of ordern ≥3 isn ?1 if the transposition graph is any unicyclic graph,that is,there aren ?1 internally disjoint trees connecting any three vertices in Un. So the upper bound forκ3(G)in Lemma 2.2 when there exist two adjacent vertices of minimum degree inGis attained.